We know that $\frac{3{{x}^{2}}}{1+{{x}^{3}}}=3{{x}^{2}}-3{{x}^{5}}+3{{x}^{8}}-...+{{\left( -1 \right)}^{n+1}}3{{x}^{3n-1}}+...$ for $x\in (-1,1)$. Using this fact, find the power series for $\ln(1+x^3)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\frac{{{x}^{3n}}}{n}}$ (Choice B) B $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\cdot 3\left( 3n-1 \right){{x}^{3n-2}}}$ (Choice C) C $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\frac{{{x}^{3n}}}{3n}}$ (Choice D) D $\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\cdot \left( 3n-1 \right){{x}^{3n-2}}}$
Explanation: First, integrate both sides of the given expression. $\int{\frac{3{{x}^{2}}}{1+{{x}^{3}}}}\text{ }dx=\int\Big[{3{{x}^{2}}-3{{x}^{5}}+3{{x}^{8}}-...+{{\left( -1 \right)}^{n+1}}3{{x}^{3n-1}}+...}\Big]dx$ Evaluate the left by using the $~u$ -sbstitution $~u=1+{{x}^{3}}\,;\text{ }du=3{{x}^{2}}dx\,$. This gives us $~\ln \left( 1+{{x}^{3}} \right)\,$. [Note that we don't need absolute values because $~-1<x<1\,$.] On the right, we obtain $ C+\left( \frac{3{{x}^{3}}}{3}-\frac{3{{x}^{6}}}{6}+\frac{3{{x}^{9}}}{9}-...+{{\left( -1 \right)}^{n+1}}\frac{3{{x}^{3n}}}{3n}+... \right)$ Thus, $\ln\left(1+x^3\right)=C+\left( \dfrac{3{{x}^{3}}}{3}-\dfrac{3{{x}^{6}}}{6}+\dfrac{3{{x}^{9}}}{9}-...+{{\left( -1 \right)}^{n+1}}\dfrac{3{{x}^{3n}}}{3n}+... \right)$ Note here that we only need to put the constant of integration on one side of the equation. If we let $~x=0\,$, we see that $~C=0~$ (because $~\ln 1=0\,$ ). Reducing each of the fractions by a factor of $~3~$ then gives the desired result. $\ln \left( 1+{{x}^{3}} \right)={{x}^{3}}-\frac{{{x}^{6}}}{2}+\frac{{{x}^{9}}}{3}-...+{{\left( -1 \right)}^{n+1}}\frac{{{x}^{3n}}}{n}+...=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\frac{{{x}^{3n}}}{n}}$